Question

Find the Laplace transform of$\mathit{f}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\underset{0}{\overset{t}{\int }}(t-v)e^{3v}dv$

Short answer

Laplace transformation of${\int }_0^t{e}^v\sin (t-v)dv$ is $\frac{1}{s^2(s-3)}$.

Step-by-step solution

Assuming g(t) and h(t).

Let

$\begin{array}{l}g\left(t\right)=t\\ h\left(t\right)=e^{3t}\end{array}$

Getting the values of G(s) and H(s).

Now,

$\begin{array}{c}G\left(s\right)=\mathcal{L}\left\{g\right(t\left)\right\}\left(s\right)\\ =\frac{1}{s^2}\end{array}$

$\begin{array}{c}H\left(s\right)=\mathcal{L}\left\{h\right(t\left)\right\}\left(s\right)\\ =\frac{1}{s-3}\end{array}$

Step 3: Now, On applying the Convolution Theorem

We get,

$\begin{array}{c}\mathcal{L}\left\{f\right(t\left)\right\}\left(s\right)=\mathcal{L}\left\{{\int }_0^t(t-v)e^{3v}dv\right\}\left(s\right)\\ =\mathcal{L}\left\{{\int }_0^{t}g(t-v)h\left(v\right)dv\right\}\end{array}$

$\begin{array}{c}\mathcal{L}\left\{f\right(t\left)\right\}\left(s\right)=\mathcal{L}\left\{g\right(t)\ast h(t\left)\right\}\left(s\right)\\ =G\left(s\right)\cdot H\left(s\right)\\ =\frac{1}{s^2(s-3)}\end{array}$

Therefore,

$\mathcal{L}\left\{f\right(t\left)\right\}\left(s\right)=\frac{1}{s^2(s-3)}$